Question

solve for x and graph the solution.
(x - 1)(x + 1) ≥ 0
plot the endpoints. select an endpoint to change it from closed to open. select the middle of a segment, ray, or line to delete it.

Explanation:

Step1: Find the critical points

Set \((x - 1)(x + 1)=0\). Using the zero - product property \(a\times b = 0\) implies \(a = 0\) or \(b = 0\). So \(x-1=0\) gives \(x = 1\), and \(x + 1=0\) gives \(x=-1\). These two points divide the number line into three intervals: \((-\infty,-1)\), \((-1,1)\), and \((1,\infty)\).

Step2: Test the intervals

• For the interval \((-\infty,-1)\), let \(x=-2\). Then \((-2 - 1)(-2 + 1)=(-3)\times(-1)=3\geq0\).
• For the interval \((-1,1)\), let \(x = 0\). Then \((0 - 1)(0 + 1)=(-1)\times(1)=-1<0\).
• For the interval \((1,\infty)\), let \(x = 2\). Then \((2 - 1)(2 + 1)=(1)\times(3)=3\geq0\).

Step3: Consider the endpoints

Since the inequality is \(\geq\), when \(x=-1\), \((-1 - 1)(-1 + 1)=0\); when \(x = 1\), \((1 - 1)(1 + 1)=0\). Both \(x=-1\) and \(x = 1\) satisfy the inequality.

Answer:

The solution of the inequality \((x - 1)(x + 1)\geq0\) is \(x\leq - 1\) or \(x\geq1\). On the number - line, we plot closed circles at \(x=-1\) and \(x = 1\), and shade the regions to the left of \(x=-1\) (including \(x=-1\)) and to the right of \(x = 1\) (including \(x = 1\)).